I want to prove what's used in the fourth line below the "Proof" section here: http://en.wikipedia.org/wiki/Quantum_relative_entropy#The_result

The statement is: Let $\rho$ be a density operator on a finite dimensional complex inner product space, with spectral decomposition $\rho = \sum_i \rho_i P_i$, with orthogonal projections $P_i$ and eigenvalues $\rho_i$. Then $$\log\rho = \sum_i P_i\log\rho_i$$

I don't see how to prove this. Of course, if I imagine the projections to be zero-matrices with a $1$ somewhere on the diagonal such that they all add up to $\mathbb{1}$, then it makes perfect sense, as then I just have the logarithm of a diagonal matrix, which is clear. But how can I see this rigorously from the properties of orthogonal projections?

  • $\begingroup$ Wikipedia's page on matrix logarithms may be helpful. Also note that it's sufficient to check that $\exp \log \rho$ is equal to $\rho$. $\endgroup$ Jul 21, 2014 at 15:36
  • $\begingroup$ That was actually quite helpful! $\endgroup$ Jul 21, 2014 at 15:55
  • $\begingroup$ Glad to hear it. Keep in mind that you're allowed/encouraged to answer your own questions, and (after a sufficient time period) will be able to accept it as the answer if you consider that the best answer. That also lets us give feedback etc in comments. $\endgroup$ Jul 21, 2014 at 15:56

1 Answer 1


Using $P_i^2 = P_i$ we get $$e^{P_i\log\rho_i} = \sum_{k=0}^\infty \frac{(P_i\log\rho_i)^k}{k!} = \mathbb{1}-P_i +P_i\sum\frac{\log^k\rho_i}{k!} = \mathbb{1}-P_i+P_i\rho_i$$ For the whole expression we find (using that $P_iP_k=\delta_{i,k}P_k$, and $\sum_iP_i=\mathbb{1}$) $$\exp \sum_iP_i\log\rho_i = \prod_i\mathbb{1}-P_i+P_i\rho_i = \mathbb{1} -\sum_iP_i + \rho = \rho.$$ Hence indeed $\log\rho = \sum_iP_i\log\rho_i$.


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